Simplify; express your answer in exponential form. Assume $k\neq 0, x\neq 0$. $\dfrac{{(k^{-2})^{3}}}{{(k^{-2}x^{5})^{-4}}}$
Solution: To start, try working on the numerator and the denominator independently. In the numerator, we have ${k^{-2}}$ to the exponent ${3}$ . Now ${-2 \times 3 = -6}$ , so ${(k^{-2})^{3} = k^{-6}}$ In the denominator, we can use the distributive property of exponents. ${(k^{-2}x^{5})^{-4} = (k^{-2})^{-4}(x^{5})^{-4}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(k^{-2})^{3}}}{{(k^{-2}x^{5})^{-4}}} = \dfrac{{k^{-6}}}{{k^{8}x^{-20}}}$ Break up the equation by variable and simplify. $\dfrac{{k^{-6}}}{{k^{8}x^{-20}}} = \dfrac{{k^{-6}}}{{k^{8}}} \cdot \dfrac{{1}}{{x^{-20}}} = k^{{-6} - {8}} \cdot x^{- {(-20)}} = k^{-14}x^{20}$.